Improper integrals

Definition and Types

Improper integrals are definite integrals that involve either:

1. Infinite limits of integration, or
2. An integrand that becomes infinite within the limits of integration

Type 1: Infinite Limits of Integration

Case 1: Integration to Infinity

Case 2: Integration from Negative Infinity

Case 3: Integration over the Entire Real Line

\(\int_{-\infty}^{\infty} f(x) \, dx = \int_{-\infty}^{a} f(x) \, dx + \int_a^{\infty} f(x) \, dx\) for any value of $a$ where $f$ is defined.

Type 2: Integrand with a Vertical Asymptote

Case 1: Asymptote at the Lower Limit

\(\int_a^{b} f(x) \, dx = \lim_{\epsilon \to 0^+} \int_{a+\epsilon}^{b} f(x) \, dx\) where $f$ has a vertical asymptote at $x = a$.

Case 2: Asymptote at the Upper Limit

\(\int_a^{b} f(x) \, dx = \lim_{\epsilon \to 0^+} \int_a^{b-\epsilon} f(x) \, dx\) where $f$ has a vertical asymptote at $x = b$.

Case 3: Asymptote at an Interior Point

\(\int_a^{b} f(x) \, dx = \int_a^{c} f(x) \, dx + \int_c^{b} f(x) \, dx = \lim_{\epsilon \to 0^+} \left[ \int_a^{c-\epsilon} f(x) \, dx + \int_{c+\epsilon}^{b} f(x) \, dx \right]\) where $f$ has a vertical asymptote at $x = c$ with $a < c < b$.

Convergence and Divergence

Definition

• An improper integral converges if the corresponding limit exists and is finite.
• An improper integral diverges if the limit doesn’t exist or is infinite.

Convergence Theorems

Comparison Test

If $0 \leq f(x) \leq g(x)$ for all $x \geq a$, then:

1. If $\int_a^{\infty} g(x) \, dx$ converges, then $\int_a^{\infty} f(x) \, dx$ also converges.
2. If $\int_a^{\infty} f(x) \, dx$ diverges, then $\int_a^{\infty} g(x) \, dx$ also diverges.

Limit Comparison Test

If $f(x) \geq 0$, $g(x) > 0$ for all $x \geq a$, and $\lim_{x \to \infty} \frac{f(x)}{g(x)} = L$ where $0 < L < \infty$, then:

• $\int_a^{\infty} f(x) \, dx$ and $\int_a^{\infty} g(x) \, dx$ both converge or both diverge.

p-Test (for Power Functions)

For $p \in \mathbb{R}$:

• $\int_1^{\infty} \frac{1}{x^p} \, dx$ converges if $p > 1$
• $\int_1^{\infty} \frac{1}{x^p} \, dx$ diverges if $p \leq 1$

For integrals with a vertical asymptote at $x = a$:

• $\int_a^{b} \frac{1}{(x-a)^p} \, dx$ converges if $p < 1$
• $\int_a^{b} \frac{1}{(x-a)^p} \, dx$ diverges if $p \geq 1$

Absolute Convergence

Conditional Convergence

Methods for Evaluating Improper Integrals

Direct Evaluation Method

1. Replace the improper integral with the appropriate limit
2. Find the antiderivative
3. Evaluate the limit

For Type 1 (Infinite Limits)

\(\int_a^{\infty} f(x) \, dx = \lim_{R \to \infty} \left[ F(R) - F(a) \right]\) where $F$ is an antiderivative of $f$.

For Type 2 (Vertical Asymptotes)

\(\int_a^{b} f(x) \, dx = \lim_{\epsilon \to 0^+} \left[ F(b) - F(a+\epsilon) \right]\) where $f$ has an asymptote at $x = a$.

Comparison Test for Non-Negative Functions

1. Find a simpler function $g(x)$ that bounds $f(x)$
2. Determine if the simpler integral $\int g(x) \, dx$ converges or diverges
3. Apply the comparison test to draw conclusions about $\int f(x) \, dx$

Asymptotic Comparison

For positive functions $f(x)$ and $g(x)$, if $f(x) \sim g(x)$ as $x \to \infty$ (meaning $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$), then:

• $\int_a^{\infty} f(x) \, dx$ and $\int_a^{\infty} g(x) \, dx$ both converge or both diverge.

Common Improper Integrals

Standard Convergent Improper Integrals

• $\int_0^{\infty} e^{-x} \, dx = 1$
• $\int_0^{\infty} e^{-ax} \, dx = \frac{1}{a}$ for $a > 0$
• $\int_1^{\infty} \frac{1}{x^p} \, dx = \frac{1}{p-1}$ for $p > 1$
• $\int_0^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}$
• $\int_0^{\infty} e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}$

Standard Divergent Improper Integrals

• $\int_1^{\infty} \frac{1}{x} \, dx$ (diverges logarithmically)
• $\int_0^1 \frac{1}{x} \, dx$ (diverges logarithmically)
• $\int_0^1 \frac{1}{x^p} \, dx$ for $p \geq 1$
• $\int_0^{\infty} \sin x \, dx$ (oscillates)
• $\int_0^{\infty} \cos x \, dx$ (oscillates)

Worked Examples

Example 1: Infinite Upper Limit

Determine if $\int_{4}^{+\infty} \frac{x+1}{x^2(x-3)} \, dx$ converges.

Solution: Step 1: Decompose into partial fractions \(\frac{x+1}{x^2(x-3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-3}\)

Solving for coefficients (by multiplying both sides by $x^2(x-3)$ and comparing): \(x+1 = Ax(x-3) + B(x-3) + Cx^2\)

Evaluating at various points:

• At $x = 0$: $1 = -3B$, so $B = -\frac{1}{3}$
• At $x = 3$: $4 = 9C$, so $C = \frac{4}{9}$

Comparing coefficients of $x^2$: $0 = A + C$, so $A = -C = -\frac{4}{9}$

Step 2: Check each term for convergence \(\int_{4}^{+\infty} \frac{x+1}{x^2(x-3)} \, dx = \int_{4}^{+\infty} \left( \frac{-4/9}{x} + \frac{-1/3}{x^2} + \frac{4/9}{x-3} \right) \, dx\)

For $\int_{4}^{+\infty} \frac{1}{x} \, dx$: This diverges by the p-test (p = 1).

Since one term diverges, the entire integral diverges.

Example 2: Convergent Integral with Infinite Limit

Establish if $\int_{1}^{+\infty} \frac{x - \arctan x}{x^3} \, dx$ converges using asymptotic comparison.

Solution: Step 1: Analyze the asymptotic behavior As $x \to \infty$, $\arctan x \to \frac{\pi}{2}$, so: \(\frac{x - \arctan x}{x^3} \sim \frac{x - \frac{\pi}{2}}{x^3} \sim \frac{x}{x^3} = \frac{1}{x^2}\)

Step 2: Apply the p-test $\int_{1}^{+\infty} \frac{1}{x^2} \, dx$ converges since $p = 2 > 1$.

Step 3: Verify using limit comparison test \(\lim_{x \to \infty} \frac{x - \arctan x}{x^3} \cdot x^2 = \lim_{x \to \infty} \frac{x - \arctan x}{x}\)

Using L’Hôpital’s rule: \(\lim_{x \to \infty} \frac{x - \arctan x}{x} = \lim_{x \to \infty} \left(1 - \frac{\arctan x}{x}\right) = 1 - \lim_{x \to \infty} \frac{\arctan x}{x}\)

Since $\lim_{x \to \infty} \arctan x = \frac{\pi}{2}$, we get: \(\lim_{x \to \infty} \frac{\arctan x}{x} = \lim_{x \to \infty} \frac{\frac{\pi}{2}}{x} = 0\)

Therefore: \(\lim_{x \to \infty} \frac{x - \arctan x}{x} = 1 - 0 = 1\)

By the limit comparison test, since this limit is finite and positive, and $\int_{1}^{+\infty} \frac{1}{x^2} \, dx$ converges, $\int_{1}^{+\infty} \frac{x - \arctan x}{x^3} \, dx$ also converges.

Example 3: Integral with Vertical Asymptote

Establish if $\int_0^1 \frac{dx}{e^{\sqrt{x}}-1}$ converges.

Solution: Step 1: Identify the behavior near the critical point As $x \to 0^+$, $\sqrt{x} \to 0$, and $e^{\sqrt{x}} \to 1$. Using Taylor series: $e^{\sqrt{x}} - 1 \approx \sqrt{x} + \frac{x}{2} + \ldots \approx \sqrt{x}$ for small $x$.

Step 2: Compare with a known function \(\frac{1}{e^{\sqrt{x}}-1} \sim \frac{1}{\sqrt{x}}\)

Step 3: Apply the p-test for vertical asymptotes $\int_0^1 \frac{1}{\sqrt{x}} \, dx = \int_0^1 x^{-1/2} \, dx$ converges since $p = 1/2 < 1$.

Step 4: Verify using limit comparison test \(\lim_{x \to 0^+} \frac{\frac{1}{e^{\sqrt{x}}-1}}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^+} \frac{\sqrt{x}}{e^{\sqrt{x}}-1}\)

Using L’Hôpital’s rule: \(\lim_{x \to 0^+} \frac{\sqrt{x}}{e^{\sqrt{x}}-1} = \lim_{x \to 0^+} \frac{\frac{1}{2}x^{-1/2}}{e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2}} = \lim_{x \to 0^+} \frac{1}{e^{\sqrt{x}}} = 1\)

By the limit comparison test, since this limit is finite and positive, and $\int_0^1 \frac{1}{\sqrt{x}} \, dx$ converges, $\int_0^1 \frac{dx}{e^{\sqrt{x}}-1}$ also converges.

Example 4: Logarithmic Integral

Establish if $\int_0^1 \ln x \, dx$ converges.

Solution: Step 1: Identify the behavior near x = 0 As $x \to 0^+$, $\ln x \to -\infty$, so this is an improper integral with a vertical asymptote at the lower limit.

Step 2: Evaluate using the definition $\int_0^1 \ln x \, dx = \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 \ln x \, dx$

Step 3: Find the antiderivative $\int \ln x \, dx = x \ln x - x + C$

Step 4: Apply the limits $\lim_{\epsilon \to 0^+} \int_{\epsilon}^1 \ln x \, dx = \lim_{\epsilon \to 0^+} [(x \ln x - x)]{\epsilon}^1$ $= \lim{\epsilon \to 0^+} [(1 \ln 1 - 1) - (\epsilon \ln \epsilon - \epsilon)]$ $= -1 - \lim_{\epsilon \to 0^+} (\epsilon \ln \epsilon - \epsilon)$

Since $\lim_{\epsilon \to 0^+} \epsilon \ln \epsilon = 0$ (which can be verified using L’Hôpital’s rule): $\int_0^1 \ln x \, dx = -1$

Therefore, the improper integral converges to -1.

Example 5: Integral with Oscillatory Behavior

Determine if $\int_1^{+\infty} \sin\left(\frac{1}{x}\right) \, dx$ converges.

Solution: Step 1: Use integration by parts Let $u = \sin\left(\frac{1}{x}\right)$ and $dv = dx$ Then $du = -\frac{1}{x^2} \cos\left(\frac{1}{x}\right) \, dx$ and $v = x$

$\int \sin\left(\frac{1}{x}\right) \, dx = x \sin\left(\frac{1}{x}\right) - \int x \cdot \left(-\frac{1}{x^2}\right) \cos\left(\frac{1}{x}\right) \, dx$ $= x \sin\left(\frac{1}{x}\right) + \int \frac{1}{x} \cos\left(\frac{1}{x}\right) \, dx$

Let $t = \frac{1}{x}$, then $dx = -\frac{1}{t^2} \, dt$ $\int \frac{1}{x} \cos\left(\frac{1}{x}\right) \, dx = \int \cos t \, dt = \sin t + C = \sin\left(\frac{1}{x}\right) + C$

Therefore: $\int \sin\left(\frac{1}{x}\right) \, dx = x \sin\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right) + C$

Step 2: Evaluate the limits $\int_1^{+\infty} \sin\left(\frac{1}{x}\right) \, dx = \lim_{R \to \infty} \left[ x \sin\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right) \right]1^{R}$ $= \lim{R \to \infty} \left[ R \sin\left(\frac{1}{R}\right) + \sin\left(\frac{1}{R}\right) - \sin(1) - \sin(1) \right]$

As $R \to \infty$, $\sin\left(\frac{1}{R}\right) \sim \frac{1}{R}$, so $R \sin\left(\frac{1}{R}\right) \to 1$ and $\sin\left(\frac{1}{R}\right) \to 0$

Therefore: $\int_1^{+\infty} \sin\left(\frac{1}{x}\right) \, dx = 1 - 2\sin(1)$

The integral converges to a finite value.

Strategy for Solving Improper Integral Problems

1. Identify the type of improper integral:
   • Infinite limit(s) of integration
   • Integrand with vertical asymptote(s)
   • Both infinity and vertical asymptote
2. For convergence analysis:
   • Use direct evaluation if the antiderivative is known and the limit can be computed
   • Apply comparison tests for functions that are difficult to integrate
   • Use asymptotic comparison for complex functions
   • Apply the p-test for power functions
3. For evaluating convergent improper integrals:
   • Replace with the appropriate limit definition
   • Find the antiderivative
   • Carefully evaluate the limit
   • If direct evaluation is difficult, consider numerical methods
4. Special cases:
   • For oscillatory integrals, consider integration by parts or special techniques
   • For rational functions, use partial fractions
   • For functions with absolute values, split the domain appropriately
5. Verification:
   • Double-check your work, especially limit evaluations
   • Ensure proper handling of indeterminate forms
   • Use alternative methods to confirm results when possible

See:

• Integration Techniques
• Antiderivatives
• L’Hôpital’s Rule
• Convergence Tests (for Series)
• Series

Exercises

1. Determine Convergence

Establish if the following integrals converge:

1. $\int_{4}^{+\infty} \frac{x+1}{x^2(x-3)} dx$

2. $\int_{1}^{+\infty} \frac{x - \arctan x}{x^3} dx$ (using asymptotic comparison with $g(x) = \frac{1}{x^2}$)

3. $\int_e^{\infty} \frac{dx}{x \ln x}$

4. $\int_0^1 \frac{dx}{e^{\sqrt{x}}-1}$

5. $\int_1^{+\infty} \sin\left(\frac{1}{x}\right) dx$

2. Compute Converging Improper Integrals

Establish if the following integrals converge and compute them:

1. $\int_0^1 \ln x dx$

2. $\int_0^1 \frac{1}{\sqrt{1-x^2}}dx$

3. $\int_0^1 \frac{\sqrt{x}}{x^2-4x}dx$

4. $\int_0^1 \frac{\ln x}{x\sqrt{x}}dx$

5. $\int_0^{+\infty} \frac{x}{1+x^2}dx$

6. $\int_{-\infty}^{+\infty} \frac{1}{1+x^2}dx$

7. $\int_1^e \frac{\ln x}{x(1 + 7 \ln x)} dx$ (compute the area beneath the graph)

3. Improper Integrability of Functions

Establish if the following functions are improperly integrable in the given interval:

1. $f(x) = xe^{1/x}$ in $(1, +\infty)$

2. $f(x) = x^2e^{-x^3}$ in $[1, +\infty)$

3. $f(x) = \frac{x-1}{x^2+1}$ in $[2, +\infty)$

4. $f(x) = \frac{1+\cos x}{\sqrt[4]{(1-x^3)^3}}$ in $[0,1]$

5. $f(x) = \frac{e^{-x}}{\sin(1/x)}$ in $[2,+\infty)$

6. $f(x) = \frac{1}{3+x^2}$ in $[1,+\infty)$

7. $f(t) = \frac{3-t}{t^2+1}$ in $[2,+\infty)$